Nathanael A. Fortune edited section_Application_to_Specific_Heat__.tex  over 8 years ago

Commit id: f00e8767c735abb6e9e2833a10528cf7c42ecc36

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\label{eq:NuclearCuContribution}  c = \alpha \frac{H^2}{T^2} \mathrm{\ [J/K]}  \end{equation}  where $\alpha = m \frac{{\lambda}_N}{{\mu}_0}$ has units $\left[\frac{\textrm{Joules Kelvin}}{(\textrm{Tesla})^2}\right]$ $\left[\frac{\textrm{J K}}{(\textrm{T})^2}\right]$  and values $\alpha_{\textrm{0.4 mg}} = 1.28 \cdot 10^{-9}$ and $\alpha_{\textrm{0.5 mg}} = 1.60 \cdot 10^{-9}$ for the Jan 2015 and July 2015 runs, respectively.